Metamath Proof Explorer


Theorem idfullsubc

Description: The source category of an inclusion functor is a full subcategory of the target category if the inclusion functor is full. Remark 4.4(2) in Adamek p. 49. See also ressffth . (Contributed by Zhi Wang, 11-Nov-2025)

Ref Expression
Hypotheses idfth.i ⊢ I = id func ⁡ C
idsubc.h ⊢ H = Hom 𝑓 ⁡ D
idfullsubc.j ⊢ J = Hom 𝑓 ⁡ E
idfullsubc.b ⊢ B = Base D
idfullsubc.c ⊢ C = Base E
Assertion idfullsubc ⊢ I ∈ D Full E → B ⊆ C ∧ J ↾ B × B = H

Proof

Step Hyp Ref Expression
1 idfth.i ⊢ I = id func ⁡ C
2 idsubc.h ⊢ H = Hom 𝑓 ⁡ D
3 idfullsubc.j ⊢ J = Hom 𝑓 ⁡ E
4 idfullsubc.b ⊢ B = Base D
5 idfullsubc.c ⊢ C = Base E
6 fullfunc ⊢ D Full E ⊆ D Func E
7 6 sseli ⊢ I ∈ D Full E → I ∈ D Func E
8 1 7 imaidfu2lem ⊢ I ∈ D Full E → 1 st ⁡ I Base D = Base D
9 4 8 eqtr4id ⊢ I ∈ D Full E → B = 1 st ⁡ I Base D
10 eqid ⊢ 1 st ⁡ I Base D = 1 st ⁡ I Base D
11 eqid ⊢ Hom ⁡ D = Hom ⁡ D
12 eqid ⊢ x ∈ 1 st ⁡ I Base D , y ∈ 1 st ⁡ I Base D ⟼ ⋃ p ∈ 1 st ⁡ I -1 x × 1 st ⁡ I -1 y 2 nd ⁡ I ⁡ p Hom ⁡ D ⁡ p = x ∈ 1 st ⁡ I Base D , y ∈ 1 st ⁡ I Base D ⟼ ⋃ p ∈ 1 st ⁡ I -1 x × 1 st ⁡ I -1 y 2 nd ⁡ I ⁡ p Hom ⁡ D ⁡ p
13 relfull ⊢ Rel ⁡ D Full E
14 1st2ndbr ⊢ Rel ⁡ D Full E ∧ I ∈ D Full E → 1 st ⁡ I D Full E 2 nd ⁡ I
15 13 14 mpan ⊢ I ∈ D Full E → 1 st ⁡ I D Full E 2 nd ⁡ I
16 10 11 12 15 5 3 imasubc ⊢ I ∈ D Full E → x ∈ 1 st ⁡ I Base D , y ∈ 1 st ⁡ I Base D ⟼ ⋃ p ∈ 1 st ⁡ I -1 x × 1 st ⁡ I -1 y 2 nd ⁡ I ⁡ p Hom ⁡ D ⁡ p Fn 1 st ⁡ I Base D × 1 st ⁡ I Base D ∧ 1 st ⁡ I Base D ⊆ C ∧ J ↾ 1 st ⁡ I Base D × 1 st ⁡ I Base D = x ∈ 1 st ⁡ I Base D , y ∈ 1 st ⁡ I Base D ⟼ ⋃ p ∈ 1 st ⁡ I -1 x × 1 st ⁡ I -1 y 2 nd ⁡ I ⁡ p Hom ⁡ D ⁡ p
17 16 simp2d ⊢ I ∈ D Full E → 1 st ⁡ I Base D ⊆ C
18 9 17 eqsstrd ⊢ I ∈ D Full E → B ⊆ C
19 16 simp3d ⊢ I ∈ D Full E → J ↾ 1 st ⁡ I Base D × 1 st ⁡ I Base D = x ∈ 1 st ⁡ I Base D , y ∈ 1 st ⁡ I Base D ⟼ ⋃ p ∈ 1 st ⁡ I -1 x × 1 st ⁡ I -1 y 2 nd ⁡ I ⁡ p Hom ⁡ D ⁡ p
20 9 sqxpeqd ⊢ I ∈ D Full E → B × B = 1 st ⁡ I Base D × 1 st ⁡ I Base D
21 20 reseq2d ⊢ I ∈ D Full E → J ↾ B × B = J ↾ 1 st ⁡ I Base D × 1 st ⁡ I Base D
22 1 7 11 2 12 8 imaidfu2 ⊢ I ∈ D Full E → H = x ∈ 1 st ⁡ I Base D , y ∈ 1 st ⁡ I Base D ⟼ ⋃ p ∈ 1 st ⁡ I -1 x × 1 st ⁡ I -1 y 2 nd ⁡ I ⁡ p Hom ⁡ D ⁡ p
23 19 21 22 3eqtr4d ⊢ I ∈ D Full E → J ↾ B × B = H
24 18 23 jca ⊢ I ∈ D Full E → B ⊆ C ∧ J ↾ B × B = H